Fractal Geometry
Isa S. Jubran – [email protected] – http://web.cortland.edu/jubrani
A
Fractal
is a geometric object whose dimension is fractional. Most fractals are self
similar, that is when any small part of a fractal is magnified the result resembles
the original fractal. Examples of fractals are the Koch curve, the Seirpinski gasket,
and the Menger sponge.
Similarity Dimension:
•
Take a line segment of length one unit and divide it into N equal parts. let
S
=
old lengthnew length
, then
N
=
S
1
.
N = 3, S = 3
•
Take a square (dimension 2) of area one square unit and divide it into N
congruent parts. Then
S
=
√
N
and
N
=
S
2.
N= 4, S= 2
•
Take a cube (dimension 3) of volume one cubic unit and divide it into
N
congruent parts. Then
S
=
N
13and
N
=
S
3.
...
... ...
... ... ...
... .
. . . . . . . . . . . . . . . . . .
... ... ... ... ... ... ... . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . ... .. .. .. .. .. .. .. .. .. .
. . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . .
. ... ............... . . . . . . . . . . . . .
... ... ... ... ... ... ...
N = 8, S= 2
So given set of dimension
d
consisting of
N
parts congruent part then
N
=
S
d,
where
S
=
new lengthold length. Solving for
d
yields
Exercise 1:
T
∞= lim
n→∞T
nis called the
Seirpinski Triangle or Gasket
.
1.
What is the perimeter of
T
n?
2.
What is the area of
T
n?
3.
Compute the fractal dimension of
T
∞.
T0 T1 T2
Solution:
1.
P
n= (
32)
nP0
.
2.
A
n= (
34)
nA0
.
3.
N
= 3 and
S
= 2, hence
d
=
ln3ln2
.
Exercise 2:
C
∞= lim
n→∞C
nis called the
Seirpinski Carpet
.
C0 C1 C2
1.
What is the perimeter of
C
n?
2.
What is the area of
C
n?
3.
Compute the fractal dimension of
C
∞.Solution:
1.
P
n=
P0
+
13P0
+
382P0
+
· · ·
+
8n
−1
3n
P0
.
2.
A
n= (
89)
nA0
.
3.
N
= 8 and
S
= 3, hence
d
=
ln8Exercise 3:
M
∞= lim
n→∞M
nis called the
Menger Sponge
.
M0 M1 M2
0 2
2
1.
What is the volume of
M
n.
2.
What is the surface area of
M
n?
3.
Compute the fractal dimension of
M
∞.
Solution:
1.
V0
= 2
3= 8 and
V
n
= 20
·
(1
/
3)
3·
(
V
n−1).
2.
A0
= 6
·
2
2= 24 and
A
n
= 20
·
(1
/
3)
2·
(
A
n−1)
−
24
·
(1
/
3)
2·
(
A
tn−1)
·
2
= (20
/
9)
·
Area
(
C
n−1)
−
(64
/
3)
·
(8
/
9)
n−1
where
Area
(
C
n−1) is the area of
the
n
−
1 stage in the construction of the Sierpinski carpet. Hence
A
n=
(8
/
9)
n·
16 + (20
/
9)
n·
8.
3.
N
= 20 and
S
= 3, hence
d
=
ln20ln3
.
Exercise 4:
Draw a fractal with dimension
32.
Solution:
... ... ... ... ... ..
... ... ... ... ... ..
✲ ✛
...
. . . . . . . . .
...
. . . . . . . . ... .. .. ..
. ...
. . . . . . . . ... .. .. ..
. ...
. . . . . . . . ... .. .. .. .
...
. . . . . . . . .
...
. . . . . . . . ... .. .. ..
. ...
. . . . . . . . ... .. .. ..
. ...
. . . . . . . . ... .. .. .. .
✻
❄
Demonstrate the applet ”Fractal Dimension” at:
Box Dimension:
Box dimension is another way to measure fractal dimension, it is defined as follows:
If
X
is a bounded subset of the Euclidean space, and
ǫ
≥
0. Let
N
(
X, ǫ
) be the
minimal number of boxes in the grid of side length
ǫ
which are required to cover
X
. We say that
X
has box dimension
D
if the following limit exists and has value
D
.
lim
ǫ→0+
ln(
N
(
X, ǫ
))
ln(
1ǫ)
Consider for example the Seirpinski gasket:
✻
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .
... ...
... ... ... ...
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .
✲
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .
...
0 1
1
ǫ
N
(
X, ǫ
)
1
1
1
2
3
1
4
9
...
...
1
2
n3
nHence D
= lim
n→∞
ln(4
·
3
n)
ln(2
n)
=
ln 3
ln 2
Demonstrate the applet ”Box Counting Dimension” at:
Geometrically, if you plot the results on a graph with ln
N
(
X, ǫ
) on the vertical
axis and ln(1
/ǫ
) on the horizontal axis, then the slope of the best fit line of the
data will be an approximation of Box dimension of the fractal.
... ... ... ... ... ...
... .. .. .. .. .. .. ... . . . . . . . . . . . . . . . . . . . .
✲ ✻
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
.. .. .. .. .. .. .. .. .. .. .
.. .. .. .. .. .. .. .. .. .. ... .. .. .. .. .. .. .. .. .. .. . . . . . . . . . . . . . . . . . . . .
(ln(1),ln(1))
(ln(4),ln(9))
(ln(2),ln(3))
✛
✛
Discuss computing the fractal dimension of a bunch of broccoli.
http://hypertextbook.com/facts/2002/broccoli.shtml
My Results:
Ball Size
N
1ǫln(
1ǫ)
ln(
N
)
6x5x2.5
10
1/4.5
-1.504
2.303
4x4x2.5
19
1/3.5
-1.253
2.94
3.5x3x2
28 1/2.83
-1.04
3.33
3x2.5x2
41
1/2.5
-.916
3.71
2.5x2.5x2 64 1/2.33
-.847
4.158
Use the linear regression applet at http://math.hws.edu/javamath/ryan/Regression.html
to plot the data points and find the slope of the best fit line.
Fractal dimension = slope = 2.62.
Divider Dimension:
A fractal curve has fractal (or divider) dimension
D
if its length
L
when measured
with rods of length
l
is given by
L
=
C
·
l
1−DC
is a constant that is a certain measure of the apparent length, and the equation
above must be true for several different values of
l
. Consider the Koch curve:
K2
l
L
1
1
1 3
4 3 1 9
16 9
...
...
1 3n
(
4 3
)
n
It appears to have length
L
= (
4 3)
n
when measured with rods of length
l
= (
1 3)
n
.
Hence,
(
4
3
)
n
=
C
·
(
1
3
)
n(1−D)
This implies that
n
ln
43
= ln(
C
) +
n
(1
−
D
) ln
1 3.
In order for the equation to be satisfied for all
n
, ln
C
must be zero, hence
C
= 1
and so
D
=
ln 4ln 3
.
Geometrically, if you graph ln(
L
) versus ln(
l
) the
D
= 1
−
Slope of line
.
Exercise:
The west coast of Britain when measured with rods of length 10
km
is
3020
km
. But when measured with rods of length 100
km
it is 1700
km
. What is
the fractal dimension of the west coast of Britain?
Demonstrate the applet ”Coastline of Hong Kong” at:
Affine transformations:
An affine transformation is of the form
f
(
Ã
x
y
!
) =
Ã
a b
c d
!
·
Ã
x
y
!
+
Ã
e
f
!
or
f
(
Ã
x
y
!
) =
Ã
r
cos(
θ
)
−
s
sin(
φ
)
r
sin(
θ
)
s
cos(
φ
)
!
·
Ã
x
y
!
+
Ã
e
f
!
Where
r
and
s
are the scaling factors in the
x
and
y
directions respectively.
θ
and
φ
measure rotation of horizontal and vertical lines resp.
e
and
f
measure horizontal
and vertical translations resp.
It can be easily shown that
r
2=
a
2+
c
2,
s
2=
b
2+
d
2,
θ
= arctan(
c/a
) and
φ
= arctan(
−
b/d
).
✲ ✛
✻
❄ θ
c
a ✛ ✲
✻
❄ b
d φ
.... .... .... .... ....
...
A′ .... .... .... .... ....
... ... ... ... ... ... ...
.... .... .... .... .... .
y′= 1 2y
. . . . . . . . . . . . . . . . . . . . .
B′ ...... ... ... ... ... ...
C
✲ ✻
A C′
x′= 1 2x Similarity
B
.... .... .... .... .... .
A B C
D A′
B′
C′.... .... .... .... ....
... ...
... ... ... ... ... ... ....
.... .... .... .... .
Af f inity
. . . . . . . . . . . . . . . . . . . .
. ...
✲ ✻
D
x′= 3 4x y′= 1
2y
A reflection about the y-axis would be given by
x
′=
−
x
and
y
′=
y
− − − − − − −
matrix f orm
Ã
−
1 0
0 1
!
A reflection about the x-axis would be given by
x
′=
x
and
y
′=
−
y
− − − − − − −
matrix f orm
Ã
1
0
0
−
1
Rotation is represented as follows:
B
B′
φ= 45
C′
y′= 1
√2y
C
.. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
θ= 0
.. .. .. .. .. .. .. .. .. .. .. .. .. ..
...
A
B
A′
y′ = 1
√2x+y
C
B′
... ...
... ...
...
θ= 45 ...
... ...
... ......
... ... ... ... ... ... ... .
A
φ= 0 ...
... ... ... ... ...
... ...
... ... ... ... ... ...
✲ ✻
... ... ... ... ... ...
...
D
... ...
... ... ... ... ... ...
✲
x′ = 1
√2x
✻
D
x′ =x− 1
√2y
M atrixf orm
:
Ã
cos(
θ
)
−
sin(
φ
)
sin(
θ
)
cos(
φ
)
!
=
Ã
a b
c d
!
... ...
... ...
...
.. .. .. .. .. .. .. .. .. .. .. .. .. ..
.. ...
.. .. .. .. .. .. .. .. .. .. .. ..
... ...
... ...
...B′
A A′
B
φ= 45
C′
y′= 1
√2x+√12y
C
θ= 45
.. .. .. .. .. .. .. .. .. .. .. .. .. ..
.. ... ... ... ... ... ...
... ✲ D
x′= 1
√2x−√12y
Shears are represented as follows:
... ... ... ... ... ...
... ✲ ✻
... ... ... ... ... ...
... .. .. .. .. .. ..
.. ...
... ... ... ... ... ... ...
... ... ... ... ...
...
... .
. . . . . . . . . . . . . . . . . . .
. ...
... ... ... ... ...
... .. .. .. .. .. .. ... . . . . . . . . . . . . . . . . . . . .
✲ ✻
... ... ... ... ... ...
... .. .. .. .. .. .. .. ...
... ... ... ... ...
... . . . . . . . . . . . . . . . . . . . .
. ...
... ... ... ... ...
... .. .. .. .. .. .. ... . . . . . . . . . . . . . . . . . . . .
A B C
D
B′
A′
x′=x
y′ = 2x+y
Shear in the y direction
A B C
D
B′
C′
x′=x+ 2y
y′=y
Iterated Function Systems:
An
Iterated Function System
in
R
2is a collection
{
F1
, F2, . . . , F
n
}
of contraction
mappings on
R
2. (
F
i
is a contraction mapping if given
x, y
then
d
(
F
i(
x
)
, F
i(
y
))
≤
sd
(
x, y
)
where
0
≤
s <
1)
The Collage Theorem
says that there is a unique nonempty compact subset
A
⊂
R
2called the attractor such that
A
=
F
1(
A
)
∪
F
2(
A
)
∪
. . .
∪
F
n(
A
)
Sketch of Proof:
Given a complete metric space
X
, let
H
(
X
) be the collection of nonempty compact
subsets of
X
. Given
A
and
B
in
H
(
X
) define
d
(
A, B
) to be the smallest number
r
such that each point of in
A
is within
r
of some point in
B
and vice versa. If
X
is complete so is
H
(
X
).
Given an IFS on
X
, define
G
:
H
(
X
)
→
H
(
X
) by
G
(
K
) =
F1
(
K
)
∪
. . .
∪
F
n(
K
). If
each
F
iis a contraction then so is
G
. The contraction mapping theorem says that
a contraction mapping on a complete space has a unique fixed point. Hence there
is
A
∈
H
(
X
) such that
G
(
A
) =
A
=
F
1(
A
)
∪
F
2(
A
)
∪
. . .
∪
F
n(
A
).
Given
K0
⊂
R
2, let
K
n
=
G
n(
K0
). Then
d
(
A, K
i)
≤
s
id
(
K0
, G
(
K0
))
1
−
s
Hence
K
iis a very good approximation of
A
for large enough
i
.
Deterministic Algorithm:
(Slow!)
Given
K0
, let
K
n=
G
n(
K0
). Now
K
n→
A
.
Random Algorithm:
(Fast!)
Choose
P0
∈
K0
and let
P
i+1=
F
j(
P
i) where
F
jis chosen at random from all of
the
F
i. Clearly
P
i∈
K
iand for sufficiently large
i
,
P
iis arbitrarily close to some
Finding IFS Rules from Images of Points
Given three non-collinear initial points
p
1 = (
x1, y1
)
, p2
= (
x2, y2
)
, p3
= (
x3, y3
)
and three image points
q1
= (
u1, v1
)
, q2
= (
u2, v2
)
, q3
= (
u3, v3
) respectively. Find
an affine transformation
T
defined by
T
(
Ã
x
y
!) =
Ãa b
c d
!·
Ãx
y
!+
Ãe
f
!such
that
T
(
p1
) =
q1, T
(
p2
) =
q2,
and
T
(
P3
) =
q3
.
This gives the following six equations in six unknowns:
ax1
+
by1
+
e
=
u1
cx
1+
dy
1+
f
=
v
1ax2
+
by2
+
e
=
u2
cx2
+
dy2
+
f
=
v2
ax3
+
by3
+
e
=
u3
cx3
+
dy3
+
f
=
v3
...... ... ... ... ... ... ... .. .. .. .. .. .. .. ... ... ... ... ... ... ... .. .. .. .. .. .. .. ... ... ... ... ... ... ... .. .. .. .. .. .. .. ... ... ... ... ... ... ... ... .. .. .. .. .. .. ... ... ... ... ... ... ... ... .. .. .. .. .. .. ... ... ... ... ... ... ... ... .. .. .. .. .. .. .. ... ... ... ... ... ... ... ... .. .. .. .. .. .. ... ... ... ... ... ... ... ... .. .. .. .. .. .. ... ... ... ... ... ... ... ... .. .. .. .. .. .. .. ... ... ... ... ... ... ... ... .. .. .. .. .. .. .. ... ... ... ... ... ... ... ... .. .. .. .. .. .. .. ... ... ... ... ... ... ... ... .. .. .. .. .. .. .. ... ... ... ... ... ... ... ... .. .. .. .. .. .. ... ... ... ... ... ... ... ... .. .. .. .. .. .. ... ... ... ... ... ... ... ... .. .. .. .. .. .. ... ... ... ... ... ... ... ... .. .. .. .. .. .. ..
.5,0,0, .5,0,0
.5,0,0, .5, .5,0 .5,0,0, .5,0, .5
The IFS for the Seirpinski triangle is
{
T1, T2, T2
}
where
T1
(
Ãx
y
!) =
Ã.
5
0
0
.
5
!
·
Ãx
y
!+
Ã0
0
!T2
(
Ãx
y
!) =
Ã.
5
0
0
.
5
!
·
Ãx
y
!+
Ã.
5
0
!T3
(
Ãx
y
!) =
Ã.
5
0
0
.
5
Example:
... ... ... ... ... ... ... .. .. .. .. .. .. ... . . . . . . . . . . . . . . . . . . . . ... ... ... ... ... ... ... .. .. .. .. .. .. ... . . . . . . . . . . . . . . . . . . . . ... ... ... ... ... ... ... .. .. .. .. .. .. ... . . . . . . . . . . . . . . . . . . . . ... ... ... ... ... ... ... .. .. .. .. .. .. ... . . . . . . . . . . . . . . . . . . . . ... ... ... ... ... ... ... .. .. .. .. .. .. ... . . . . . . . . . . . . . . . . . . . . ... ... ... ... ... ... ... .. .. .. .. .. .. ... . . . . . . . . . . . . . . . . . . . . ... ... ... ... ... ... ... .. .. .. .. .. .. ... . . . . . . . . . . . . . . . . . . . . ... ... ... ... ... ... ... .. .. .. .. .. .. ... . . . . . . . . . . . . . . . . . . . . ... ... ... ... ... ... ... .. .. .. .. .. .. ... . . . . . . . . . . . . . . . . . . . . ✲ ✻I II III
IV V
V I V II V III
0 0 1 1 2 3 1 3 2 3 1 3
The IFS for the Seirpinski carpet is
{
T1, T2, T3, T4, T5, T6, T7, T8
}
where
T1
(
Ãx
y
!) =
Ã.
333
0
0
.
333
!
·
Ãx
y
!+
Ã0
0
!T2
(
Ãx
y
!) =
Ã..
333
0
0
.
333
!
·
Ãx
y
!+
Ã.
333
0
!T3
(
Ãx
y
!) =
Ã.
333
0
0
.
333
!
·
Ãx
y
!+
Ã.
666
0
!T
4(
Ã
x
y
!) =
Ã.
333
0
0
.
333
!
·
Ãx
y
!+
Ã0
.
333
!T5
(
Ãx
y
!) =
Ã.
333
0
0
.
333
!
·
Ãx
y
!+
Ã.
666
.
333
!T6
(
Ãx
y
!) =
Ã.
333
0
0
.
333
!
·
Ãx
y
!+
Ã0
.
666
!T7
(
Ãx
y
!) =
Ã.
333
0
0
.
333
!
·
Ãx
y
!+
Ã.
333
.
666
!T8
(
Ãx
y
!) =
Ã.
333
0
0
.
333
Example:
.5,−.5,5, .5, .5,5 .5, .5,5,−.5, .5,15
−10< x <30 0< y <40
(20,15)
T runk 0,0,10,0, .333,0 (15,20) (5,20)
(0,5) (5,15) (15,15)
(10,0) (10,10)
. . . . . . . . . . . ... ... ... .. .. .. .. .. . ... ... ... .. ... .. .. .. .. .. . .. .. .. .. .. . ... ... ... .. . . . . . . . . . . ... .. .. .. .. . .... .. .. .. . ... ... ... .... . . . . . . . . . ... .. .. .. . .. .. .. .. .. . .. ... ... ... .... . . . . . . . . . .... .. .. .. . ... . ... ... .. .. .. .. .. . ... ... ... ... .. .. .. .. . . . . . . . . . . ... .. .. .. .. . ... ... ... .... . . . . . . . . . .... .. .. .. . ... . ... ... .. .. .. .. .. . ... ... ... ... .. .. .. .. . . . . . . . . . . ... .. .. .. .. . ... ... ... .... . . . . . . . . . .... .. .. .. . .. . . . . . . . . . . . ... ... ... .. .. .. .. .. . ... ... ... .. ... .. .. .. .. .. . .. .. .. .. .. . ... ... ... .. . . . . . . . . . . ... .. .. .. .. . .... .. .. .. . ... ... ... .... . . . . . . . . . ... .. .. .. . .. .. .. .. .. . .. ... ... ... .... . . . . . . . . . .... .. .. .. . ... . ... ... .. .. .. .. .. . ... ... ... ... .. .. .. .. . . . . . . . . . . ... .. .. .. .. . ... ... ... .... . . . . . . . . . .... .. .. .. . ... . ... ... .. .. .. .. .. . ... ... ... ... .. .. .. .. . . . . . . . . . . ... .. .. .. .. . ... ... ... .... . . . . . . . . . .... .. .. .. . .. . . . . . . . . . . . ... ... ... .. .. .. .. .. . ... ... ... .. ... .. .. .. .. .. . .. .. .. .. .. . ... ... ... .. . . . . . . . . . . ... .. .. .. .. . .... .. .. .. . ... ... ... .... . . . . . . . . . ... .. .. .. . .. .. .. .. .. . .. ... ... ... .... . . . . . . . . . .... .. .. .. . ... . ... ... .. .. .. .. .. . ... ... ... ... .. .. .. .. . . . . . . . . . . ... .. .. .. .. . ... ... ... .... . . . . . . . . . .... .. .. .. . ... . ... ... .. .. .. .. .. . ... ... ... ... .. .. .. .. . . . . . . . . . . ... .. .. .. .. . ... ... ... .... . . . . . . . . . .... .. .. .. . .. . . . . . . . . . . . ... ... ... .. .. .. .. .. . ... ... ... .. ... .. .. .. .. .. . .. .. .. .. .. . ... ... ... .. . . . . . . . . . . ... .. .. .. .. . .... .. .. .. . ... ... ... .... . . . . . . . . . ... .. .. .. . .. .. .. .. .. . .. ... ... ... .... . . . . . . . . . .... .. .. .. . ... . ... ... .. .. .. .. .. . ... ... ... ... .. .. .. .. . . . . . . . . . . ... .. .. .. .. . ... ... ... .... . . . . . . . . . .... .. .. .. . ... . ... ... .. .. .. .. .. . ... ... ... ... .. .. .. .. . . . . . . . . . . ... .. .. .. .. . ... ... ... .... . . . . . . . . . .... .. .. .. . .. ✲ ✻
The IFS for the tree is
{
T1, T2, T2
}
where
T1
(
Ãx
y
!) =
Ã.
5
.
5
−
.
5
.
5
!
·
Ãx
y
!+
Ã5
15
!T2
(
Ãx
y
!) =
Ã.
5
−
.
5
.
5
.
5
!
·
Ãx
y
!+
Ã5
5
!T3
(
Ãx
y
!) =
Ã0
0
0
.
333
!
·
Ãx
y
!+
Ã10
0
!Download and experiment with the freeware ”IFS Construction Kit” which can be
used to design and draw fractals based on iterated function systems.
You can find it at:
http://ecademy.agnesscott.edu/
∼
lriddle/ifskit/
•
Discuss the IFS for the Fern.
•
Discuss the IFS for the Tree.
•
Discuss the IFS for the mountain range.
Mandelbrot and Julia Sets:
Consider the complex function
F
c(
z
) =
z
2+
c
. Given an initial point
z0
define its
orbit under
F
cto be
z0, z1, z2, . . .
where
z
n+1=
z
n2+
c
.
Define the escape set,
E
c, and the prisoner set,
P
c, for for the parameter
c
to be:
E
c=
{
z0
:
|
z
n| → ∞
as n
→ ∞}
,
and
E
c=
{
z
0|
z
0∈
/
E
c}
Now the Julia set for the parameter
c
,
J
c, is the boundary of the escape set
E
c.
For example, if
c
= 0 then
F
c(
z
) =
z
2. Now if
z
=
re
iθthen
F
(
z
) =
r
2e
2iθ.
|
z
0|
θ
|
z
0|
θ
|
z
0|
θ
z
.8
15
1
15
1.5
15
z
2.64
30
1
30
2.25
30
z
4.4096
60
1
60
5.0625
60
z
8.1678 120
1
120
25.629
120
z
16
.0281 240
1
240
656.84
240
z
32.0008 120
1
120
431439.88 120
✲ ✻
❄
✛ ✲
✻
❄ ✛
z0 z1 z2
z3 z4
|z0|<1 |z0|= 1
z0 z1 z2 z3
z4
The Julia set is either connected (one piece) or totally disconnected(dust).
Now we define the Mandelbrot set to be
M
=
{
c
∈
C
:
J
cis connected
}
Experiment
with
the
Java
applet
”Mandelbrot
and
Julia
Sets”
at:
http://classes.yale.edu/fractals/
Download ”winfeed” at the address below and use it to plot the Julia sets for the
constants
c
=
−
.
5 +
.
5
i
,
c
=
−
.
7454285 + 0
.
1130089
i
, and others.
